The monthly revenue of a certain company is given by R=800p-7p^2, where p is the price in dollars of the product that the company manufactures. At what price will the revenue be $10,000, if the price is to be greater than $50?

1. Start with the revenue formula:

R = 800p - 7p^2

2. Plug $10,000 into the revenue formula:

10000 = 800p - 7p^2

3. Rearrange the equation to standard quadratic form:

7p^2 - 800p + 10000 = 0

4. Solve for \( p \) using the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 7 \), \( b = -800 \), and \( c = 10000 \).

p = \frac{800 \pm \sqrt{(-800)^2 - 4 \cdot 7 \cdot 10000}}{2 \cdot 7}

p = \frac{800 \pm \sqrt{640000 - 280000}}{14}

p = \frac{800 \pm \sqrt{360000}}{14}

p = \frac{800 \pm 600}{14}

5. Calculate the two possible solutions for \( p \):

p_1 = \frac{800 + 600}{14} = \frac{1400}{14} = 100

p_2 = \frac{800 - 600}{14} = \frac{200}{14} \approx 14.29

6. Since the price must be greater than $50:

p = 100